Relative lowering of vapour pressure

A colligative property depends only on the number of solute particles dissolved, never on their chemical identity — equal moles of sugar or urea in the same solvent produce exactly the same effect. Relative lowering of vapour pressure is the simplest of these, and the one from which the syllabus builds outward to the other three.

Setting up Raoult's law. For a solution of a non-volatile solute (subscript 2) in a volatile solvent (subscript 1), Raoult's law states that the solvent's partial vapour pressure over the solution is proportional to its own mole fraction:

where: $p_1$ is the vapour pressure of the solution, $p_1^0$ is the vapour pressure of the pure solvent at the same temperature, and $x_1$ is the mole fraction of solvent in the solution. Because the solute is non-volatile, it contributes essentially nothing to the vapour above the liquid — every molecule that escapes is a solvent molecule, and fewer of them do, since some surface sites are now occupied by solute particles that never leave.

Turning proportion into lowering. Since the solution has only two components, $x_1 + x_2 = 1$, so $x_1 = 1-x_2$. Substituting:

Rearranging gives the lowering itself, and dividing by $p_1^0$ makes it relative:

where: $p_1^0-p_1$ is the absolute lowering of vapour pressure (always positive) and $x_2$ is the mole fraction of solute. The left side is dimensionless — a pure ratio — and it equals the solute's mole fraction exactly, with no approximation so far. That equality is why the quantity is colligative: it reads off the proportion of solute molecules among all molecules present, and says nothing about whether those molecules are large or small, sugar or salt-like. Below, an SVG contrasts the surface of a pure solvent, where every surface site can host an escaping molecule, against a solution, where solute particles block a fraction of those sites — the visual reason fewer molecules escape and vapour pressure drops.

Weighing an unknown solute. If $w_2$ g of solute (molar mass $M_2$) dissolves in $w_1$ g of solvent (molar mass $M_1$), and the solution is dilute enough that $n_2 \ll n_1$, the exact mole fraction $x_2 = n_2/(n_1+n_2)$ simplifies to $x_2 \approx n_2/n_1$. Writing moles as mass over molar mass:

Every quantity here except $M_2$ is measurable in a lab, so this single relation converts a vapour-pressure measurement into a molar mass — the classic use of the topic and the reason it is examined so heavily.

Why it anchors the whole colligative-properties chapter. Freezing-point depression, boiling-point elevation and osmotic pressure all trace back to the same cause: a non-volatile solute lowers the solvent's vapour pressure at every temperature, which in turn shifts the temperatures at which the solution freezes and boils. Once this first derivation is secure, the remaining three formulae read as consequences rather than facts to memorise separately.

18 g of glucose ($M=180\ \text{g mol}^{-1}$) is dissolved in 180 g of water ($M=18\ \text{g mol}^{-1}$) at a temperature where pure water's vapour pressure is $17.5\ \text{mm Hg}$. Find (a) the vapour pressure of the solution, and (b) if instead this solution's vapour pressure is measured as $17.33\ \text{mm Hg}$ and used to check against a second, unknown solute experiment where 5 g of solute in 100 g of benzene ($M=78\ \text{g mol}^{-1}$) drops benzene's vapour pressure from 640 to 631 mm Hg, find the unknown solute's molar mass.

Step 1. Moles in the glucose solution: $n_{glucose} = 18/180 = 0.1\ \text{mol}$, $n_{water} = 180/18 = 10\ \text{mol}$.

Step 2. Mole fraction of solute: $x_2 = \dfrac{0.1}{0.1+10} = \dfrac{0.1}{10.1} = 0.0099$.

Step 3. Apply the relation: $p_1^0 - p_1 = 17.5 \times 0.0099 = 0.173\ \text{mm Hg}$, so $p_1 = 17.5 - 0.173 = \mathbf{17.33\ mm\ Hg}$ — matching the value quoted, confirming the setup.

Step 4. For the benzene experiment, relative lowering: $\dfrac{p_1^0-p_1}{p_1^0} = \dfrac{640-631}{640} = \dfrac{9}{640} = 0.01406$.

Step 5. Use the dilute-solution molar-mass formula with $w_2=5$ g, $M_1=78\ \text{g mol}^{-1}$, $w_1=100$ g: $0.01406 = \dfrac{5\times78}{M_2\times100}$, giving $M_2 = \dfrac{390}{1.406} = \mathbf{277.4\ g\ mol^{-1}}$.

Competitive stretch

If the same benzene experiment used $150$ g of benzene instead of $100$ g, keeping $w_2=5$ g and the same relative lowering $0.01406$, the formula gives $M_2 = \dfrac{5\times78\times1.5}{100\times0.01406\times1}$ — more directly, $M_2 = \dfrac{w_2 M_1}{(\Delta p/p_1^0)\, w_1}$ scales linearly with $w_1$, so tripling $w_1$ while holding the lowering fixed would demand a proportionally larger $M_2$; examiners use this to test whether students recompute from the formula rather than pattern-match the first answer.

• Forgetting that the dilute-solution approximation $x_2 \approx n_2/n_1$ needs $n_2 \ll n_1$; for concentrated solutions, use the exact $x_2 = n_2/(n_1+n_2)$.
• Swapping $p_1^0$ and $p_1$ in the ratio — the numerator is always (pure minus solution), a positive lowering, never the reverse.
• Applying the simple relation to a volatile solute, where the solute itself adds vapour pressure and the plain Raoult's-law form no longer holds.
• Forgetting the solute must be non-dissociating; if it ionises or associates, a van't Hoff factor $i$ must be introduced (covered separately under abnormal molar mass).

Ask the tutor: want to try a few more molar-mass-from-vapour-pressure problems, including ones where the exact (non-approximated) mole fraction is required?